Game with Pearls（二分图）

Game with Pearls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 145    Accepted Submission(s): 96

Problem Description

Tom and Jerry are playing a game with tubes and pearls. The rule of the game is:

1) Tom and Jerry come up together with a number K.

2) Tom provides N tubes. Within each tube, there are several pearls. The number of pearls in each tube is at least 1 and at most N.

3) Jerry puts some more pearls into each tube. The number of pearls put into each tube has to be either 0 or a positive multiple of K. After that Jerry organizes these tubes in the order that the first tube has exact one pearl, the 2nd tube has exact 2 pearls, …, the Nth tube has exact N pearls.

4) If Jerry succeeds, he wins the game, otherwise Tom wins.

Write a program to determine who wins the game according to a given N, K and initial number of pearls in each tube. If Tom wins the game, output “Tom”, otherwise, output “Jerry”.

 

 

Input

The first line contains an integer M (M<=500), then M games follow. For each game, the first line contains 2 integers, N and K (1 <= N <= 100, 1 <= K <= N), and the second line contains N integers presenting the number of pearls in each tube.

 

 

Output

For each game, output a line containing either “Tom” or “Jerry”.

 

 

Sample Input

2

5 1

1 2 3 4 5

6 2

1 2 3 4 5 5

 

 

Sample Output

 

Jerry

Tom

 

     题意：

     给出 T 组样例，后给出 N 和 K，代表有 N 条管子，后给出 N 条管子上面的初始状态珍珠数。可以向任意一条管子添加 K 的珍珠数，问能否使管子分别出现 1 ~ N 的数量。

 

     思路：

     二分图，N 最大值才100，所以建图也是非常方便的。将每个数 ans 加 K 循环至 ans > N 来建图，最后判断二分匹配后是否满配即可。

 

     AC：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAX = 105;

int n, k;
int num[MAX];

int G[MAX][MAX];
int linker[MAX];
bool vis[MAX];

void build () {
    memset(G, 0, sizeof(G));

    for (int i = 1; i <= n; ++i) {
        for (int j = num[i]; j <= n; j += k) {
            G[i][j] = 1;
        }
    }
}

bool dfs (int x) {
    for (int i = 1; i <= n; ++i) {
        if (G[x][i] && !vis[i]) {
            vis[i] = 1;
            if (linker[i] == -1 || dfs(linker[i])) {
                linker[i] = x;
                return true;
            }
        }
    }

    return false;
}

int hungary () {
    int res = 0;
    memset(linker, -1, sizeof(linker));

    for (int i = 1; i <= n; ++i) {
        memset(vis, 0, sizeof(vis));
        if (dfs(i)) ++res;
    }

    return res;
}

int main () {

    int t;
    scanf("%d", &t);

    while (t--) {

        scanf("%d%d", &n, &k);

        for (int i = 1; i <= n; ++i) {
            scanf("%d", &num[i]);
        }

        build();

        if (n == hungary()) printf("Jerry\n");
        else printf("Tom\n");

    }

    return 0;
}